deemeetar

Less than g

Oh thank you. I've imagined it a lot more different. So here's my solution: [math]sin(45deg) = \frac{Fg2}{Fg}[/math] [math]Fg2= \frac{\sqrt{2} * Fg}{2}[/math] [math]m * a = \frac{\sqrt{2} * m * g}{2} [/math] [math]a = \frac{\sqrt{2} * g}{2} [/math]

Well my imaginary triangle was from the perpendicular(horizontal) and the parallel force(vertical). I guess since the move has different direction that the acceleration will be different but how much different. I know the perpendicular force is 0, the parallel one is G = mg. I just can't make a connection to the one under angle of 45 deg. Please tell me how to calculate that? What should i use? Am i close with calculating it like i was trying or i am far from it?

I still can't think of a way to calculate that. Maybe use Pythagorean theorem .. but how exactly idk. Maybe since the acceleration i'm interested in is the hypotenuse of the triangle... [math]x = \sqrt{( sin(45 deg) * g ) ^2 + (cos(45 deg) * 0) ^2 }[/math] [math]x = \sqrt{\frac{1}{2} * g^2 + 0 }[/math] [math]x = \frac{1}{\sqrt{2}} * g[/math] Am i close?

Well there is, but the no friction part makes me confused.

Lets say there is an object on a ramp of 45 degrees. There is no friction between the object and the ramp. Will the object have the same acceleration as an object in free falling?(g)